3. Whether an accident scene is safe or not, you should first take action to help any victims.

What is your Favorite Diesel Truck?

jok3333 [9.3K]

Answer:

Race truck go vroom

Explanation:

Truck go fast

3 0

3 years ago

Read 2 more answers

Air enters a compressor at 100 kPa and 25 ⁰C. It is compressed to 2 MPa and exits the compressor at 540 K. The compressor is at

AysviL [449]

Answer:

(a) The reversible work is 207 kJ/kg

(b) The irreversibility rate is -38.39 kJ/kg

Explanation:

State1 : p1 = 100kpa, T1= 25+273 =298k

From air table, h1 =298.18 kJ/kg, s10= 1.69528 kJ/kgK

State 2a:p2=2mpa,t2=540k (actual condition 2a)

h2a= 544.35 kJ/kg,s2a0=2.29906

actual work input to the compressor =wout=h1-h2+Qin

=298.18-544.35+(-150)kJ/kg(- sign indicate heat loss)

=(-246.17)kJ/kg(-ve sign indicates the work is given into the system

a) Reversible work= Win actual - any irreversiblities present

=246.17 + irreversibilty

b) irreversibility = T0(Entopy generation Sgen) for air, Sgen

=s20-s10-Rln(p2/p1), T0=250C

=(25+273)(s2a0-s10-Rlnp2/p1+Qout/Tsurr)

= 298x[(2.29906-1.69528-0.287kJ/kgK xln(2000kpa/100) + 150 /298]

= -38.39 kJ/kg

a)Reversible work = Win actual -any irreversiblities present

=246.17 + irreversibilty

=246.17+-38.39

=207 kJ/kg

8 0

3 years ago

Find the dy/dx using implicit differentiation

irina [24]

Answer:

dy/dx = (1 − 2x + 8y) / (4 + 3x − 12y)

Explanation:

d/dx (x − 4y) = d/dx (e^(2x + 3y − 1))

1 − 4 dy/dx = e^(2x + 3y − 1) (2 + 3 dy/dx)

Since x − 4y = e^(2x + 3y − 1):

1 − 4 dy/dx = (x − 4y) (2 + 3 dy/dx)

1 − 4 dy/dx = 2 (x − 4y) + 3 (x − 4y) dy/dx

1 − 4 dy/dx = 2x − 8y + (3x − 12y) dy/dx

1 − 2x + 8y = (4 + 3x − 12y) dy/dx

dy/dx = (1 − 2x + 8y) / (4 + 3x − 12y)

6 0

3 years ago

An electric current of 237.0 mA flows for 8.0 minutes. Calculate the amount of electric charge transported. Be sure your answer

aleksley [76]

Answer:

amount of electric charge transported = 1.13 × $10^{-2}$ C

Explanation:

given data

electric current = 237.0 mA = 0.237 A

time = 8 min = 8 × 60 sec = 480 sec

solution

we get here amount of electric charge transported that is express as

amount of electric charge transported = electric current × time ...........1

put here value and we get

amount of electric charge transported = 0.237 × 480

amount of electric charge transported = 113.76 C

amount of electric charge transported = 1.13 × $10^{-2}$ C

4 0

3 years ago

A rigid tank of volume of 0.06 m^3 initially contains a saturated mixture of liquid and vapor of H2O at a pressure of 15 bar and

Serga [27]

Answer:

The total mass in the tank = 0.45524 kg

The amount of heat transferred = 3426.33 kJ

Explanation:

Given that:

The volume of the tank V = 0.06 m³

The pressure of the liquid and the vapor of H2O (p) = 15 bar

The initial quality of the mixture $\mathbf{x_{initial} - 0.20}$

By applying the energy rate balance equation;

$\dfrac{dU}{dt} = Q_{CV} - m_eh_e$

where;

$m_e =- \dfrac{dm_{CV}}{dt}$

Thus, $\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e$

If we integrate both sides; we have:

$\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}$

$m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)$

We obtain the following data from the saturated water pressure tables, at p = 15 bar.

Since:

$h_e =h_g$

Then: $h_g = h_e = 2792.2 \ kJ/kg$

$v_f = 1.1539 \times 10^{-3} \ m^3 /kg$

$v_g = 0.1318 \ m^3/kg$

Hence;

$v_1 = v_f + x_{initial} ( v_g-v_f)$

$v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )$

$v_1 = 0.02728 \ m^3/kg$

Similarly; we obtained the data for $u_f \ \& \ u_g$ from water pressure tables at p = 15 bar

$u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg$

Hence;

$u_1 = u_f + x_{initial } (u_g -u_f)$

$u_1 =843.16 + 0.2 (2594.5 -843.16)$

$u_1 = 1193.428$

However; the initial mass $m_1$ can be calculated by using the formula:

$m_1 = \dfrac{V}{v_1}$

$m_1 = \dfrac{0.06}{0.02728}$

$m_1 = 2.1994 \ kg$

From the question, given that the final quality; $x_2 = 1$

$v_2 = v_f + x_{final } (v_g - v_f)$

$v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})$

$v_2 = 0.1318 \ m^3/kg$

Also;

$u_2 = u_f + x_{final} (u_g - u_f)$

$u_2 = 843.16 + 1 (2594.5 - 843.16)$

$u_2 = 2594.5 \ kJ/kg$

Then the final mass can be calculated by using the formula:

$m_2 = \dfrac{V}{v_2}$

$m_2 = \dfrac{0.06}{0.1318}$

$m_2 = 0.45524 \ kg$

Thus; the total mass in the tank = 0.45524 kg

FInally; from the previous equation (1) above:

$m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)$

$Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)$

Q = [(0.45524)(2594.5) -(2.1994)(1193.428)-(2792.2)(0.45524-2.1994)]

Q = [ 1181.12018 - 2624.825543 - (2792.2)(-1.74416 )]

Q = 3426.33 kJ

Thus, the amount of heat transferred = 3426.33 kJ

7 0

3 years ago